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Bài 5 hình 1: (tự vẽ hình nhé bạn)
a) Xét ΔABD và ΔACB ta có:
\(\widehat{BAD}\)= \(\widehat{BAC}\) (góc chung)
\(\widehat{ABD}\)= \(\widehat{ACB}\) (gt)
=> ΔABD ~ ΔACB (g-g)
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CB}\) = \(\dfrac{AD}{AB}\) (tsđd)
b) Ta có: \(\dfrac{AB}{AC}\) = \(\dfrac{AD}{AB}\) (cm a)
=> \(AB^2\) = AD.AC
=> \(2^2\) = AD.4
=> AD = 1 (cm)
Ta có: AC = AD + DC (D thuộc AC)
=> 4 = 1 + DC
=> DC = 3 (cm)
c) Xét ΔABH và ΔADE ta có:
\(\widehat{AHB}\) = \(\widehat{AED}\) (=\(90^0\))
\(\widehat{ADB}\) = \(\widehat{ABH}\) (ΔABD ~ ΔACB)
=> ΔABH ~ ΔADE
=> \(\dfrac{AB}{AD}\) = \(\dfrac{AH}{AE}\) = \(\dfrac{BH}{DE}\) (tsdd)
Ta có: \(\dfrac{S_{ABH}}{S_{ADE}}\) = \(\left(\dfrac{AB}{AD}\right)^2\)= \(\left(\dfrac{2}{1}\right)^2\)= 4
=> đpcm
Tiếp bài 5 hình 2 (tự vẽ hình)
a) Xét ΔABC vuông tại A ta có:
\(BC^2\) = \(AB^2\) + \(AC^2\)
\(BC^2\) = \(21^2\) + \(28^2\)
BC = 35 (cm)
b) Xét ΔABC và ΔHBA ta có:
\(\widehat{BAC}\) = \(\widehat{AHB}\) ( =\(90^0\))
\(\widehat{ABC}\) = \(\widehat{ABH}\) (góc chung)
=> ΔABC ~ ΔHBA (g-g)
=> \(\dfrac{AB}{BH}\) = \(\dfrac{BC}{AB}\) (tsdd)
=> \(AB^2\) = BH.BC
=> \(21^2\) = 35.BH
=> BH = 12,6 (cm)
c) Xét ΔABC ta có:
BD là đường p/g (gt)
=> \(\dfrac{AD}{DC}\) = \(\dfrac{AB}{BC}\) (t/c đường p/g)
Xét ΔABH ta có:
BE là đường p/g (gt)
=> \(\dfrac{HE}{AE}\) = \(\dfrac{BH}{AB}\) (t/c đường p/g)
Mà: \(\dfrac{AB}{BC}\) = \(\dfrac{BH}{AB}\) (cm b)
=> đpcm
d) Ta có: \(\left\{{}\begin{matrix}\widehat{HBE}+\widehat{BEH}=90^0\\\widehat{ABD}+\widehat{ADB=90^0}\\\widehat{HBE}=\widehat{ABD}\end{matrix}\right.\)
=> \(\widehat{BEH}=\widehat{ADB}\)
Mà \(\widehat{BEH}=\widehat{AED}\) (2 góc dd)
Nên \(\widehat{ADB}=\widehat{AED}\)
=> đpcm
Em kiểm tra lại đề bài nhé \(\frac{2}{x-y}\)hay \(\frac{2}{x-2}\)
\(1,\\ a,\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\\ 2,\\ a,\Leftrightarrow\left(x+5\right)^2=0\Leftrightarrow x=-5\\ b,\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ c,\Leftrightarrow\left(x-9\right)^2=0\Leftrightarrow x=9\\ d,\Leftrightarrow\left(x-3\right)^3=0\Leftrightarrow x=3\\ e,\Leftrightarrow3x\left(x^2-2x+3\right)=0\\ \Leftrightarrow3x\left(x^2-2x+1+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2+2=0\left(vô.nghiệm\right)\end{matrix}\right.\\ \Leftrightarrow x=0\)
\(f,\Leftrightarrow3x\left(x^2-4x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Bài 1:
a) \(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
b) \(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-7\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow\left(x+5\right)^2=0\Rightarrow x=-5\)
b) \(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
c) \(\Rightarrow\left(x-9\right)^2=0\Rightarrow x=9\)
d) \(\Rightarrow\left(x-3\right)^3=0\Rightarrow x=3\)
e) \(\Rightarrow3x\left(x^2-6x+9\right)=0\)
\(\Rightarrow3x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
f) \(\Rightarrow3x\left(x^2-4x+4\right)=0\)
\(\Rightarrow3x\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
(oh) hóa trị 1 mà zn hóa trị 2=> cthh la zn(oh)2
với lại ko có oh2 dau chi co OH hoac la H2O